I was reading Herrlich & Strecker's Category Theory, and there is a theorem called The Canonical construction of Pullbacks which states that if a category has products and equalizers, then it has pullbacks.

So I started to ask myself for some kind of converse. For instance, it is false that pullbacks and equalizers imply products, since the category of fields has pullbacks and equalizers but not products.

Then I asked if products and pullbacks imply equalizers. I haven't been able to come up with a counterexample, so I guess it must be true. Is there a proof of this?

2 Answers
2

Yes, products and pullbacks imply equalizers. The equalizer of $f,g: A \to B$ is the pullback of $(1,f)$ and $(1,g): A \to A\times B$. A cone $C$ over the pullback diagram of $(1,f)$ and $(1,g)$ has maps $\pi_1,\pi_2: C \to A$ and by commutativity of the diagram we know $\pi_1 = \pi_2$ and $f \circ \pi_1 = g \circ \pi_2$. This is clearly equivalent to a cone over the equalizer diagram of $f,g: A \to B$.

Since I deleted my (wrong) answer, let me just add an important comment here. In categories with a final element, $1$, a product is just a pullback of $X\to 1$ and $Y\to 1$. So in categories with pullbacks and a final element, there are both products and equalizers.
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Thomas AndrewsFeb 19 '13 at 20:41

Marlu's construction tells you that equalisers can be constructed as long as you have binary products and pullbacks of split monomorphisms along split monomorphisms, which is an extremely weak assumption. Let me give an alternative construction which is sometimes useful in geometry:

The proof is straightforward enough. Here what I assume is that we have binary products and pullbacks of split monomorphisms along arbitrary morphisms. The reason why I prefer this construction is that it says something extra: in $\textbf{Top}$, for example, it says that the equaliser of $f$ and $g$ is a closed subspace of $X$ if $Y$ is Hausdorff. (Similarly, in $\textbf{Sch}$, the equaliser of $f$ and $g$ is a closed subscheme if $Y$ is separated.)